Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 9 - Inferences from Two Samples - 9-3 Two Means: Independent Samples - Basic Skills and Concepts - Page 465: 17


a) Reject the null hypothesis. b)$\mu_1-\mu_2$ is between -5.9287 and 6.57372.

Work Step by Step

a) Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1$ is more than $\mu_2$. Hence the value of the test statistic: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(87.22727-86.90476)-(0)}{\sqrt{14.29263^2/22+8.988352^2/21}}=0.089.$ The degree of freedom: $min(n_1-1,n_2-1)=min(22-1,21-1)=20.$ The corresponding P-value by using the table: p is less than 0.005. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P is less than $\alpha=0.05$, because it is less than 0.005, hence we reject the null hypothesis. b) The corresponding critical value using the table: $t_{\alpha/2}=t_{0.05}=1.725.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=1.725\sqrt{\frac{14.29263^2}{22}+\frac{8.988352^2}{21}}=6.25121.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(87.22727-86.90476)-6.25121=-5.9287 and $\overline{x_1}-\overline{x_2}+E$=(87.22727-86.90476)+6.25121=6.57372.
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